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    Hello everyone, let’s start lecture 14.And our topic remains same, on the basis of mixed potential theory.This explanation of corrosion events on the basis of mixed potential theory that is thetopic and today we will consider some numerical analysisfor two situations.One is when cathodic increases, remain same and second thing isincreases but.So these two situations.We will solve some problems okay.Now if we try to solve the first case.And remember the data will take in such a fashion that they will be related, both theproblems will be both the cases will be related.Now if we try to draw the diagram, in fact here we have to drawa log I capital I.Remember here we have to consider in the form of current in order to understand the areaeffect and our choice of solution of course hydrogen evolution is our reduction reaction,pH 3 and so if pH is 3 so this E equilibrium hydrogen equal to minus 0.059 into pH equalto 05 and pH is nothing but 3.So I can multiply by 3 which is becoming minus 0.774 and our consideration is the pressureof hydrogen remains one atmosphere.Now if we arbitrarily choose i zero hydrogen over zinc surface equal to 10 to the powerof minus 11 ampere per centimeter square.i zero zinc over zinc equal to 10 to the power of minus 9 ampere per centimeter square.Now, i corr of zinc when zinc equal to one centimeter square equal to.Anyhow when you are talking about small I It’s every time it is with reference toone centimeter square unit area is 10 to the power of minus 5 ampere per centimeter squarethen i zero hydrogen on platinum equal to 10 to the power of minus 5 ampere per centimetersquare.beta a equal to 0.1 as well as beta c the only thing is their signs will change.In case of beta c there will be minus and in case of beta a there will be positive soI am just writing plus/minus volt per decade of current density and then and these twovalues will not change.When we converted it into log current scale then that means this becomes volt per decadeof current change.Now if we try to plot them so this becomes i zero hydrogen over zinc and this becomesi zero zinc over zinc so this is 9 so this point will be relative position will be likethis.And then, this point becomes my i zero hydrogen over platinum.Now if we try to seeso this is i corr zinc when it is, in the immersed in that acid without connecting itto the platinum area platinum.And if we connect it with the platinum so it will also have its own plot and then wehave to add as per the mix potential theory we have to add it.This is the total i am not indicating that because you please go back and then checkhow we have drawn this.So now this will be extended so this point becomes my E mix zinc platinum one centimetersquare area.Now once we changed it to 100 centimeters area so that say if I change it to 100 centimetersquare in case of platinum.See one case platinum equal to 100 centimeter square another case it become and zinc onecentimeter square this case one.Case two platinum equal to one centimeter square and zinc 100 centimeter square okay.So, or let’s say I make it different little different so I make it 10 centimeter squareokay.So once we do that the case one case so if it is if this value is 10 to the power minus5 ampere per centimeter square so this point would become 10 to the power of minus 3 amperewhen platinum equal to 100 centimeter square.Please make sure that please look at the unit what we are using so it is here it is ampere.Now when you do that we have its own plot.This is Ic total in ampere equal to Ic hydrogen over sink of 1 centimeter square plus Ic hydrogenover platinum 100 centimeter square.Sorry this is not that part, it is rather this is the this is Ic hydrogen over platinum100 centimeter square and then if I do it with the different color so this becomes mythis is the total part.Now if we extend this one so it will cut here this becomes my E mix zinc platinum 100 centimeterssquare and zinc here we are not changing.Now those values if we try to indicate this is 10 to the power of minus 5 ampere per centimetersquare and this is 10 to the power of minus 11 ampere per centimeter square this is 10to the power of minus 9 ampere per centimeter square and we have already indicated everythingand that this value corresponds to minus 0.177 volt.Now we have to find out E mix zinc.So now if you want to mix I try to find out so the first thing let’s find out E mixzinc.In order to do that we just take this line if we take that line so E mix minus E, sothis is nothing but equilibrium hydrogen, this is E equilibrium zinc, equilibrium hydrogenequal to minus 0.1 log of ic hydrogen over zinc divided by i zero hydrogen over zinc.Now if we solve it by putting all the values we will getE mix would be equal to minus 0.6 minus 0.177 equal to minus 0.1777.This will be the E mix zinc.Now if we try to solve, now we have to find out E equilibriums zinc which can also befind out found out we have to find out that part.So E mix zinc minus E equilibrium zinc equal to 0.1 log of we have to take this particularline ia zinc i zero zinc over zinc surface would beequal to E equilibrium zinc would be equal tominus 0.4 minus 0.777 equal to minus 1.177.So that will be the value, Okay.So we have found out this point so i can mention it to be minus 0.7.So these points i can change it becomes minus 1.177 and this would bethis would be minus 0.777 okay.So now if we try to find out now let’s find out, this point.This point if i need to find out so let’s find out E mixzinc platinum when it is one centimeter square, if we try to find out we have to solve twoequations.We have to find out this point.So this point would become 10 to the power of minus 11 plus 10 to the power of minus5 which is nothing but 10 to the power of minus 5 ampere per centimeter square.If we solve that so two equations if I indicate in this fashion E mix single prime.So single prime minus 0.177 plus equal to minus 0.1 log ic total by i zero total.I can write in this fashion 1 log of ia divided by i zero total, the way we have done previouslyso it becomes so this becomes a mix prime plus 0.177 equal to minus 0.1 log of ia plus.And then this value is nothing but 10 to the power hydrogen total is nothing but 10 tothe power of minus 5 ampere per centimeter square, we have seen already so this becomes0.5 sorry this becomes 0.5 and for the anodic part if we consider plus 1.177 equal tominus 0.0 this is plus log ia plus 0.9.Because we are considering this line and this line.So if we solve that this is equation 1 this is equation 2.So it if we add them prime plus 1.354 equal to these two would get cancelled and thenwe are getting 0.4.So E mix prime becomes 477.So this becomes this value becomes minus 0.477 volt.Now we can calculate ia zinc when for this couple of one centimeter square we can dothat so if we take equation 2 so minus 0.477 plus 1.7177 equal to 0.1 log ia by 10 to thepower minus 9.Because i zero here I have to take i zero of zinc over zinc surface.So if we solve this this is for solve this will get ia would be equal to 10 to the power7 into 10 to the power minus 9.So this becomes 10 to the power minus 2 ampere per centimeter square.So this is the corrosion rate of zinc when is galvanically coupled zinc one centimetersquare is galvanically coupled to platinum one centimeter square.Now when we increase the area of the platinum so we have to find out so if we correspondto this particular point because this so this becomes 10 to the power minus 3 because weare not changing i zero capital I zero because zinc area we are not changing so 10 to thepower minus 11 ampere plus 10 to the power minus 3 ampere.Which is roughly equal to 10 to the power minus 3 ampere because minus 11 because whenwe change everything to ampere instead of ampere per second ampere per centimeter squareand then this particular value we can calculate.The similar practice we can do.So if we try to find out E mix zinc 1 centimeter square platinum 100 centimeter square area.So we have to find out if we assume this is to be E mix.double prime, so again we can follow the similar equation solving mode so E mix double primeplus 0.177 equal to minus 0.1 log.Now we are changing everything into capital I. Minus 0.3 because here equation becomes0.1 log Ic hydrogen I would say total plus I zero hydrogen total so this is equal toc.So I can change it to minus 0.1 log Ia divided by I zero hydrogen total so this I zero hydrogentotal is nothing but 10 to the power minus 3 ampere.So this is one equation then another equation if we consider the anodic apart 0.1 log Iaminus this will become plus.And since the zink area is not changing so i zero small i zero nothing but the capitalI zero in terms of magnitude.So if we solve this again if we add them so 2 E double prime mix plus 1.354 for this twowill get cancelled.So we are left with 0.6.So E double prime mix equal to minus 0.377 volt, 377 volt.Now this is I am talking about this point this is the value we have calculated.Now if we want to calculate Ia this is 5 this is 6, Ia when zinc one centimeter square isconnected to platinum 100 centimeter square.If we want to calculate so E mix double prime plus 1.77 equal to 0.1 Ia zero zinc over zincsurface.So this becomes 10 to the power minus 1.If we do that you will get another minus one because thisis 10 to the power minus 1.So this becomes my corrosion current when zinc 1 centimeter square is connected to platinum100 centimeter square.Now if I try to calculate the current density the current density, or corrosion currentdensity of zinc so this will be equal to 10 to the power minus 1 ampere 1 centimeter squarebecause the zinc area has not changed so this becomes the 1 ampere per centimeter square.So now you could see that, when if we try to compare this value and this value I couldsee that the corrosion rate has increased one order magnitude.Now this is the part 1 or case 2, this is case one case two if we try to see where zincarea is 10 centimeter square and platinum area is 1 centimeter square that case if weplot everything in the red color so this point will not change for one centimeter squarearea but once we change to zinc is changed to 10 centimeter square.So this way will come here.So this becomes 10 to the power minus 8 ampere since zinc is now 10 centimeter square sothis is to be 10 to the power minus 9 into 10 becomes this and this will also become10 to the power minus 10 ampere.So if we continue this so they will connect here at the same so here it will connect.Now this line we have to consider addition of this line and this line we have to considerso this will shift little bit on the right side.And we have to continue that 10 centimeter square line to this.So now we are getting new point this and so let’s find out all those two points andtry to find out what could be the corresponding Ia and ia.These two values let’s find out because this will be the mix valuesmix potential as per the the mix potential this the theory will be met at this pointat this point it will be met.So let’s find out that.Now in order to find out first we have to find out this particular point and then wego ahead with this.So if I try to find out the first point so now you will be considering everything interms of i capital I and E of course so now first part is we have to solve in order toget this point we have to solve two equations what are those two equation I will just abit of jump I’ll do first part is E mix star, let’s say this point isE mix star.So if I try to find out this E mix star minus 0.177 because this value will not change thisvalue will remain fixed volt this is volt equal to 0.1 if we take this line.So it will be in terms of area in terms of current Ic hydrogen over zinc surface I zerohydrogen over zinc surface zinc.So I can replace it with Ia log Ia and this one I note in 10 so this become 1.0 and correspondingE mix star plus 1.177 equal to 0.1 log Ia and this becomes plus 0.8.So if we add them this will get cancelled so this become 0.2.So E mix star 2 equal to minus 1.554 so E mix star equal to minus 0.777.So interestingly we could See that even if we increase the area this E mix is nothingbut the same value.Now we have to find out Ia which can be found out by taking this equation so Ia if you calculateyou will see that Ia value comes out to be 10 to the power minus 4 ampere.Now corresponding, ia if I try to find out 10 to the power of minus 4/10 centimeter squareequal to 10 to the power minus 5 ampere per centimeter square and so this is the corrosionrate of zinc of 10 centimeter square when zinc only is dipped in acid.Now we have to find out then this particular point which let’s say make it double star.So if we do that then this is let’s say 1 then 2 same I will go by I will try to solvein this case we solve this line and this line and now in this case will solve this lineand this line now we have already found out that this point will bethis point will be 10 to the power minus 5 ampere onlyBecause if we add this value and I zero when platinum is 1 centimeter square which is thispoint if we correspondingly I zero with the same value because we are not changing platinumarea.So if we add this and quantity which becomes to minus 5 ampere.So that case we have to again constitute two equations in order to do that.So E mix double star so if I try to solve this line and this line then I have to againconstitute another equation this is stage 2, E mix we are trying to find out E mix andIa first I will try to find out Ia when zinc 10 centimeter square is connected to platinum1 centimeter square.So if I try to do that so E mix plus 0.177 equal to minus 0.1 I’m not because whenwe tried to the same way when we try to solve draw a equation.An equation for this line this line so we are assuming that a current value does notchange because it is roughly same value as we have seen that this quantity and this quantityif we add it becomes roughly and under the minus 5 so that does not change.So and at the same time we are seeing that at this point rate the current correspondingto the total evolution of hydrogen is equal to the total current corresponding to thedissolution of zinc.So if we do that log Ia and here I zero total is nothing but minus 5 so it becomes so Izero total equal to 10 to the power minus 10 plus 10 to the power minus 5 equal to 10to the power minus 5 ampere so 0.5 E mix plus 1.177 equal to 0.1.This is here I am replacing Ic total so here it will be this is in the anodic part thisis the cathodic part plus 0.8.Because here I have to consider this line so here the point has 10 to the power minus8.So if we add them plus 1.354 equal to this two would get cancelled so this becomes 0.3.So E mix double star equal to minus 1.054 by 2 equal to minus 0.527.To this is the E mix thing then we can take this equation to find out Ia.So 527 if we take if we try to find out Ia.So we can solve equation 2.This equation we can solve so minus 0.527 plus 1.177 equal to 0.1 log of Ia dividedby 10 to the power minus 8 ampere so then it becomes Ia equal to 10 to the power of6.5 into 10 to the power of minus 8 which becomes 10 to the power of 1.5 minus 1.5 ampere.So now this is the corrosion current when zinc is 10 centimeter square.So in a corresponding Ia if we try to find out 10 to the power of minus 1.5 divided by10 centimeter square which becomes 10 to the power of minus 2.5 ampere per centimeter square.This is the corrosion density.When 10 centimeter square is connected to platinum one centimeter square.Now let’s get back to the corrosion current density when zinc one centimeter square isconnected to platinum one centimeter square.That value was we have calculated before so that value was 10 to the power of minus 2.That value was let me see what was the value calculated, yes this was the value we havecalculated.So it was 10 to the power of minus 2.So I could see that Ia zinc one centimeter square Platinum one centimeter square is lessis greater than Ia zinc ten centimeter square and platinum one centimeter square.So it is very clear that the corrosion current density of zinc, if we increase the anodicarea decreases with reference to the value what we have found out when zinc of one centimetersquare area is connected to platinum upon one centimeter square area.So the zinc corrosion rate actually decreases because of increase in anodic area but ifI try to find out corrosion current density when zinc one centimeter square and platinum100 centimeter square.These were the situation that time corrosion current density was ampere to the minus oneampere per centimeter square which is much higher so that means I could see that thecorrosion rate of anodic region increases because of connection of that anodic regionto a cathodic region of higher area.But the zinc corrosion rate decreases when it is connected to when its own area increaseswith reference to the cathodic area and remember every time we are assuming that it is uniformdissolution.That situation and the assumption is this is one important assumption all those problemsit is uniform distribution of cathodic and anodic reactions over cathode and anode areaand we are considering uniform dissolution.So that case we could see that increase in cathodic areas need not good for the structuraldesign rather if we have increase in anodic area with reference to the cathodic area thatsolves better because the dissolution rate decreases.So from this numerical solving, the problem solving mode we could see the effect of areafactor in a area factor in a better fashion and appreciate it better.Let’s stop it and we will continue our discussion in our future lectures, thank you.